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(A) Let the required line be parallel to the vector $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$.The position vector of the point $(1, 2, -4)$ is

Vedclass · Accepted Answer

$\vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})$

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(A) Let the required line be parallel to the vector $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$.The position vector of the point $(1, 2, -4)$ is $\vec{a} = \hat{i} + 2\hat{j} - 4\hat{k}$.The equation of the line passing through $\vec{a}$ and parallel to $\vec{b}$ is $\vec{r} = \vec{a} + \lambda\vec{b}$.The given lines are parallel to vectors $\vec{v}_1 = 3\hat{i} - 16\hat{j} + 7\hat{k}$ and $\vec{v}_2 = 3\hat{i} + 8\hat{j} - 5\hat{k}$.Since the required line is perpendicular to both lines,its direction vector $\vec{b}$ is given by the cross product $\vec{v}_1 	imes \vec{v}_2$.$\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -16 & 7 \ 3 & 8 & -5 \end{vmatrix} = \hat{i}(80 - 56) - \hat{j}(-15 - 21) + \hat{k}(24 + 48) = 24\hat{i} + 36\hat{j} + 72\hat{k}$.Dividing by $12$,we get the direction ratios as $(2, 3, 6)$.Thus,$\vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k}$.The vector equation is $\vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})$.

Find the vector equation of the line passing through the point $(1, 2, -4)$ and perpendicular to the two lines: $\frac{x-8}{3} = \frac{y+19}{-16} = \frac{z-10}{7}$ and $\frac{x-15}{3} = \frac{y-29}{8} = \frac{z-5}{-5}$.

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